Question

(i) Write Biot-Savart's law in vector form.
(ii) Two identical circular coils A and B, each of radius \( R \), carrying currents \( I \) and \( \sqrt{3} I \) respectively, are placed concentrically in XY and YZ planes respectively. Find the magnitude and direction of the net magnetic field at their common centre.

Hint:

The magnetic field at the center of a circular loop is along its axis, and for perpendicular loops, the net field is the vector sum of the components, often at an angle determined by their magnitudes.

Answer 1

Part (a)(i): Biot-Savart Law in Vector Form
The Biot–Savart law gives the magnetic field $ \vec{dB} $ at a point due to a small current element $ \vec{dl} $ carrying current $ I $, at a distance $ \vec{r} $ from the element: \[ \vec{dB} = \frac{\mu_0}{4\pi} \frac{I (\vec{dl} \times \hat{r})}{r^2} \quad \text{or equivalently} \quad \vec{dB} = \frac{\mu_0}{4\pi} \frac{I (\vec{dl} \times \vec{r})}{r^3} \] where $ \mu_0 $ is the permeability of free space, $ r = |\vec{r}| $, and $ \hat{r} = \frac{\vec{r}}{r} $.

Part (a)(ii): Net Magnetic Field at the Common Centre
Step 1: Describe the setup.
- Coil A: In the XY-plane, radius $ R $, current $ I $, center at (0, 0, 0).
- Coil B: In the YZ-plane, radius $ R $, current $ \sqrt{3} I $, center at (0, 0, 0). 

Step 2: Magnetic field due to Coil A:
Since Coil A lies in the XY-plane, its axis is along the Z-axis. The magnetic field at the center is: \[ B_A = \frac{\mu_0 I}{2R} \quad \Rightarrow \quad \vec{B_A} = \frac{\mu_0 I}{2R} \hat{k} \] (assuming counterclockwise current viewed from $ +\hat{k} $) 

Step 3: Magnetic field due to Coil B:
Coil B lies in the YZ-plane, so its axis is along the X-axis. The magnetic field at the center is: \[ B_B = \frac{\mu_0 \sqrt{3} I}{2R} \quad \Rightarrow \quad \vec{B_B} = \sqrt{3} \frac{\mu_0 I}{2R} \hat{i} \] (assuming counterclockwise current viewed from $ +\hat{i} $) 

Step 4: Net magnetic field:
\[ \vec{B}_{\text{net}} = \vec{B_B} + \vec{B_A} = \left( \sqrt{3} \frac{\mu_0 I}{2R} \right) \hat{i} + \left( \frac{\mu_0 I}{2R} \right) \hat{k} \] Magnitude: \[ B_{\text{net}} = \sqrt{ \left( \sqrt{3} \frac{\mu_0 I}{2R} \right)^2 + \left( \frac{\mu_0 I}{2R} \right)^2 } = \sqrt{(3 + 1)} \cdot \frac{\mu_0 I}{2R} = \frac{\mu_0 I}{R} \] Direction: \[ \tan \theta = \frac{B_A}{B_B} = \frac{1}{\sqrt{3}} \Rightarrow \theta = 30^\circ \] The field is in the XZ-plane, at $30^\circ$ from the $+\hat{i}$ axis toward $+\hat{k}$. 

Final Answer:
- (i) Biot–Savart Law: \[ \vec{dB} = \frac{\mu_0}{4\pi} \frac{I (\vec{dl} \times \vec{r})}{r^3} \] - (ii) Net magnetic field: magnitude = $ \frac{\mu_0 I}{R} $, direction = in XZ-plane, $30^\circ$ from X-axis toward Z-axis.